Answer
$$G'\left( y \right) = {y^{\sin y}}\left( {\frac{{\sin y}}{y} + \ln y\cos y} \right)$$
Work Step by Step
$$\eqalign{
& G\left( y \right) = {y^{\sin y}} \cr
& {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr
& G\left( y \right) = {e^{\ln {y^{\sin y}}}} \cr
& G\left( y \right) = {e^{\sin y\ln y}} \cr
& {\text{evaluate the derivative}} \cr
& G'\left( y \right) = \frac{d}{{dy}}\left( {{e^{\sin y\ln y}}} \right) \cr
& {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr
& G'\left( y \right) = {e^{\sin y\ln y}}\frac{d}{{dy}}\left( {\sin y\ln y} \right) \cr
& {\text{product rule}} \cr
& G'\left( y \right) = {e^{\sin y\ln y}}\left( {\frac{{\sin y}}{y} + \ln y\cos y} \right) \cr
& {\text{simplify}} \cr
& G'\left( y \right) = {y^{\sin y}}\left( {\frac{{\sin y}}{y} + \ln y\cos y} \right) \cr} $$
